## Friday, May 13, 2022

### An Introduction To Differential Geometry (Part 4)

Arc  length and Curvature:

In our treatment of differential geometry,  arc length  is given by:

L = Öå 3 =1 ( y i  -  x i ) 2

Where ( x i , y i) define the end points of a curve segment. Hence, when defining the length of an arc C of a curve we can technically approximate it using a sequential series of broken lines or chords, as depicted below:

The length of such broken line can easily be determined if the end points of each chord are known. Let i  then be a series of broken lines or chords whose end points lie on C.  Then if i   tends to some limit s,  C is said to be rectifiable and is called the length of the arc C.  Let x(t) then (a <  b) be an allowable representation of an arc of a curve of class r > 1.  Then the arc C has length:

s = ò a    Ö {å 3 =1  (dx i /dt) 2 } dt

And s is independent of the choice of allowable representation. Note that if we replace the fixed value b in the above with the variable t then s becomes a function of t. Note also that a can be replaced with any other fixed value. Thereby we obtain the integral:

s(t) =  ò a     Ö(x' · x') dt

The function s(t) is then called the arc length of C. If  t > t then s(t) is positive and equal to the arc a(o) b (t)  of C. If t< o  then s(t) is negative and the length of a(o) b (t)  is given by - s(t). Instead of:

s2 =  å 3 =1 i 2   =  x' · x'

We can write:  ds2 = å 3 =1 dx i 2   d· dx

where  ds is called the element of arc.  For example, looking at the circular helix we find:

x'  =  (- r sin t, r cos t, c),  x' · x'  =  r2 + c 2

So that:  s(t)  =  tÖ r2 + c 2

Then if we also wish to represent the circular helix by a parametric representation with arc length s as parameter, we can write:

x(s) =  (r cos (s/w), r sin (s/w), cs/w )  And:

w=   Ö r2 + c 2

x 1 = t,

x 2 = t sin 1/t   {t   0
0  {t = 0

x 3 = 0   (0 <  t   <  t 1

We note the function is continuous even at t = 0, but the corresponding point set - which doesn't form an arc according to the definition, has no length.

Example (3): If the arc length in polar coordinates can be obtained from, e.g.:
and  p/2then the arc length between  p/2  and  p/2 for the Archimedean spiral:
r  =  q - sin  q, and sketch the curve.

Soln.  By integration from the polar eqn.

We will need to perform:

r(q) 2   = (q - sin  q) 2

=  (q - sin  q) (q - sin  q) =

q 2  - 2 q sin  q +  sin  q

And:   d r(q)  / d q   =  1 -  cos  q

(d r(q)  / d q ) 2   =

(1 - cos  q) (1 - cos  q)  =  1  - 2 cos  q  +  cos 2  q

Whence:

r(q) 2  +   (d r(q)  / d q ) 2   =

q 2  - 2 q sin  q +  sin  q  +   1  - 2 cos  q  +  cos 2  q

Leaving the integral:

ò p / 2 0 Ö{( q 2  - 2 q sin  q +  sin 2  q)  + ( 1  -  2 cos  q  +  cos 2  q)} d q

=>   ò p / 2 0 Ö{q 2  -  2 cos  q - 2 q sin  q  +2 } d q

(Rem:  sin 2 q  +  cos 2  q  =  1)

And the full curve sketched :

The second integral is analogous - simply changing the limits - whereupon we find:

Suggested Problems:

1)A curve is given in spherical coordinates i  by:

x 1 = t,    x 2 = arcsin 1/t,    x 3  =  (t 2 – 1) 1/2

Compute the length of the arc between t = 1 and t = 2

2) Obtain the arc length s of the curve:

f(x) =  x3 / 2   -   x2 /

Between x= 0 and x=  2

3) Determine the arc length of a catenary with parametric representation:  x(t) =  (t,  a cosh (t/a), 0)

4) Find the full arc length of the Archimedian spiral shown by changing the integral to the correct limits.